Actual formulas for trigonometric functions? (no calculator)

What I need is a function, like that of a calculator, for example cosine, which only takes the radian value of angle and returns a value, just like the calculator, but I need to manually calculate it. Those are formulas/procedures I need but I have never been able to find.

From what I can see, for example Cosine, would take the angle in Radians, and would asume a Radius of 1. Since Cosine of an angle is equals to Rx/R, and since Radius is assumed to be 1 and the angle is known, only Rx would need to be determined (presumably would be the direct value Cosine would return), maybe with Pythagoras, but I can't see how if Pythagoras formulas are supposed to use Sine, Cosine, etc., as well. Then I need that you tell me the actual procedure for manually finding Sine, Cosine, and the rest of functions.

If you can also, provide a rationale as to what the different functions are doing. If you can't provide a very extensive explanation, the formulas alone would suffice and I would try to figure them out myself.

If you can, please tell me also where I can find a full, in-depth analysis of the unit circle, like Pythagoras would relate and develop the different components, with the formulas I asked for above, and why they are being developed a particular way, so I can study further by myself, and any other web resource and/or book I should refer to make sense of this normally canned-only functions.

Any information will be welcome, as I see the need of being able to calculate these trigonometric functions manually, without using the calculator or tools, to greatly and really improve math understanding.

What I need is a function, like that of a calculator, for example cosine, which only takes the radian value of angle and returns a value, just like the calculator, but I need to manually calculate it. Those are formulas/procedures I need but I have never been able to find.

From what I can see, for example Cosine, would take the angle in Radians, and would asume a Radius of 1. Since Cosine of an angle is equals to Rx/R, and since Radius is assumed to be 1 and the angle is known, only Rx would need to be determined (presumably would be the direct value Cosine would return), maybe with Pythagoras, but I can't see how if Pythagoras formulas are supposed to use Sine, Cosine, etc., as well. Then I need that you tell me the actual procedure for manually finding Sine, Cosine, and the rest of functions.

If you can also, provide a rationale as to what the different functions are doing. If you can't provide a very extensive explanation, the formulas alone would suffice and I would try to figure them out myself.

If you can, please tell me also where I can find a full, in-depth analysis of the unit circle, like Pythagoras would relate and develop the different components, with the formulas I asked for above, and why they are being developed a particular way, so I can study further by myself, and any other web resource and/or book I should refer to make sense of this normally canned-only functions.

Any information will be welcome, as I see the need of being able to calculate these trigonometric functions manually, without using the calculator or tools, to greatly and really improve [your] math understanding.

Unless you have studied power series I don't see how the functions you're asking for can possbly be of any use to "really improve math understanding."

For most angles, there are no simple formulas. A few can be done simply. For example, it is easy to see that cos(0)= 1, sin(0)= 0 and then tan(0)= 0/1= 0, cot(0)= 1/0 is undefined, sec(0)= 1/0 is undefined, csc(0)= 1/1= 1.

For pi/4, consider a right triangle with one acute angle pi/4. Since pi/4+ pi/4= pi/2, the other angle is also pi/4. This is an isosceles triangle- the two legs are of the same length. If we take the length of those legs to be 1, the length of the hypotenuse is \sqrt{1^2+ 1^2}= \sqrt{2}. Then sin(pi/4)= 1/\sqrt{2}= \sqrt{2}/2, cos(pi/4)= 1/\sqrt{2}= \sqrt{2}/2, tan(pi/4)= 1/1=1, tan(pi/4)= 1/1= 1, sec(pi/4)= \sqrt{2}/1= \sqrt{2}, csc(pi/4)= \sqrt{2}/1= \sqrt{2}.

For pi/3, consider an equilateral triangle. It has all three angles equal and there sum must be pi. Each is pi/3. If you drop a line from one vertex perpendicular to the opposite side, it is easy to show, geometrically, that it bisects that side. If we take the side of the equilateral triangle to be 1, we have two right triangles, with hypotenuse of length 1, side adjacent to the pi/3 angle of length 1/2 and so the other leg is of length \sqrt{1- (1/2)^2}= \sqrt{1- 1/4}=\sqrt{3/4}= \sqrt{3}/2. Now we have sin(pi/3)= (\sqrt{3}/2)/1)= \sqrt{3}/2, cos(pi/3)= (1/2)/1= 1/2,tan(pi/3)= (\sqrt{3}/2)/(1/2)= \sqrt{3}, cot(pi/3)= (1/2)/(\sqrt{3}/2)= 1/\sqrt{3}= \sqrt{3}/3, sec(pi/3)= 1/(1/2)= 2, csc(pi/3)= 1/(\sqrt{3}/2}= 2/\sqrt{3}= 2\sqrt{3}/3.

pi/6+ pi/3= pi/2 so pi/6 is the complement to pi/2 and we can just swap the functions and their "co" functions above. sin(pi/6)= cos(pi/3)= 1/2, cos(pi/6)= sin(pi/3)= \sqrt{3}/2, tan(pi/6)= cot(pi/3)=\sqrt{3}/3,sec(pi/6)= csc(pi/3)= 2\sqrt{3}/3, csc(pi/6)= sec(pi/3)= 2.